# Thread: Inverse of function WITH a limit??

1. ## Inverse of function WITH a limit??

Find f^-1(x) <-- inverse if f(x) = x^2 - 4x + 3, x <= 2 find value of f^-1(8).

I got x=29 BUT x<= 2????

2. Originally Posted by mwok
Find f^-1(x) <-- inverse if f(x) = x^2 - 4x + 3, x <= 2 find value of f^-1(8).

I got x=29 BUT x<= 2????
What did you find your inverse function to be?

Is it $\displaystyle f^{-1}(x)=\pm\sqrt{x+1}+2$?

3. Originally Posted by masters
What did you find your inverse function to be?

Is it $\displaystyle f^{-1}(x)=\pm\sqrt{x+1}+2$?
I just replaced x with y so my inverse function became

inverse(x) = y^2 - 4y + 3

which yield 29....is that correct? can't be cause x should be less than 2 (or equal to)

4. Originally Posted by mwok
I just replaced x with y so my inverse function became

inverse(x) = y^2 - 4y + 3

which yield 29....is that correct? can't be cause x should be less than 2 (or equal to)
Not quite. Your original function was $\displaystyle f(x)=x^2-4x+3$

$\displaystyle f(x)=y=x^2-4x+3$

Interchange x and y

$\displaystyle x=y^2-4y+3$

Isolate the y

$\displaystyle y^2-4y=x-3$

Complete the square on the left side. You'll see why in a minute.

$\displaystyle y^2-4y+4=x-3+4$

$\displaystyle (y-2)^2=x+1$

$\displaystyle y-2=\pm\sqrt{x+1}$

$\displaystyle y=\pm\sqrt{x+1}+2$

$\displaystyle \boxed{f^{-1}(x)=\pm\sqrt{x+1}+2}$

5. Wow! I didn't know you could plus 4 to each side and solve the problem like that. Are there multiple ways to solve this?

6. Originally Posted by mwok
Wow! I didn't know you could plus 4 to each side and solve the problem like that. Are there multiple ways to solve this?
You could've used the quadratic formula, but completing the square seems to be the best way.

Now, can you find $\displaystyle f^{-1}(8)$ ?

7. Originally Posted by masters
You could've used the quadratic formula, but completing the square seems to be the best way.

Now, can you find $\displaystyle f^{-1}(8)$ ?
Yep, got it. It can either be -1 or 5 BUT since x should be less or equal to 2, it is automatically -1.

Thanks so much!